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Since 'a' is a prime number, without loss of generality we can consider it to be an odd number (the exception is 2, and it is clear that 2*2+26 is not a prime).Then 'a' can be written as "a = 2b+1", for some 'b' a natural number.a*a = (2b+1)*(2b+1) = 4b^2 + 4b + 1.(a*a)+26 = 4b^2 + 4b + 1 + 26 = (4b*(b+1))+27 ...... (*)

But to show that (a*a)+26 is not prime, it suffices to show that (4b*(b+1))+27 is a multiple of three. Of course 27 is a multiple of three, so we focus on the second term. But then either 'b' or 'b+1' must be a multiple of three, which is equivalent to say that 'b-1' (OR 'b+2') is not a multiple of three. This is because, for any natural number X, one and only one of 'X', 'X+1', 'X+2' is divisible by three.

a = 2b+1==> b = (a-1)/2==> b - 1 = ((a-1)/2) - 1 = (a-3)/2 .... (**)In order 'b-1' to be a multiple of three, 'a' must be a multiple of three -- which cannot be true because 'a' is prime. Therefore, either 'b' or 'b+1' must be a multiple of three, which implies (4b*(b+1))+27 is a multiple of three. And hence (a*a)+26 is a not prime.

If anyone noticed a flaw with the argument, it would be nice to post your comments.

I can't immediately spot the problem with your proof, but it must be flawed, because your same reasoning can apply to a*a + 8, which fails at a=3.

I forgot to include the condition that 'a' should be greater than 3 so that it can be written in the form of '2b+1' for 'b' a natural number. This is because 'b-1' will be zero if you take a=3. And as we all know zero is a multiple of any number.

Even in the original puzzle, if you take a=3, a*a+26 will be 35 which is not a prime number but it is not a multiple of three either. In a sense you can say my argument fails. Thanks for pointing me the error.

It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met:x > 0 (just to be safe, this may not be necessary),(x+1)%3 = 0 (e.g. 27 from the original)2*2+x is not prime (since this case was proved by example)3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1)

This appears sound to me right now, but it's starting to get late. Let me know if it checks out.

It would appear from this that 4b(b+1)+1+x would give you a working case back to (a^2)+x if the following conditions are met:x > 0 (just to be safe, this may not be necessary),(x+1)%3 = 0 (e.g. 27 from the original)2*2+x is not prime (since this case was proved by example)3*3+x is not prime (this is needed because if (b-1)%3=0, then a%3=0; thus for a=2b+1 when (b-1)%3=0, a is not prime unless it is 3. This allows us to state that 4b(b+1) is divisible by 3 and therefore not prime as long as b > 1)

This appears sound to me right now, but it's starting to get late. Let me know if it checks out.